# Curvature of a special curve

In the following problem I am asked to find the curvature of a curve. The problem is the following: \

Given the curve $$\alpha(t), \alpha: \mathbb{R} \rightarrow \mathbb{R}^3$$, which is parametrised by arc-length, we define the curve $$\beta(t) = \alpha'(t)$$. Show that the curvature of $$\beta,\kappa_{\beta} = \sqrt{1+\frac{\tau^2}{\kappa_{\alpha}^²}}$$, where $$\tau$$ is the torsion of $$\alpha$$ and $$\kappa_{\alpha}$$ its curvature.

My attempt is the following:
I know that $$\kappa_{\beta} = \frac{||\beta' \times \beta''||}{||\beta^3||}, ||\beta' \times \beta''|| = (||\beta'^2||||\beta''^2|| - (\beta'\cdot\beta'')^2)^\frac{1}{2}$$. However, since torsion of $$\alpha$$ is $$\tau = N'\cdot B$$, I don't know could the torsion appear here. I also noticed that $$N = T_{\beta}$$, where $$N$$ is the normal vector of $$\alpha$$ and $$T_{\beta}$$ the tangent unit vector of $$\beta$$. I would be grateful if somebody could give me a hint in order to solve the exercise.

• Actually write out the Frenet equations. Commented Mar 5 at 20:24
• Where would it lead it to me? Commented Mar 5 at 20:27
• To the correct answer. Expand the cross-product explicitly in terms of the Frenet frame (of $\alpha$). By the way, classically, $\beta$ is called the tangent indicatrix of $\alpha$. Commented Mar 5 at 20:28

Let me write out the details for this one. We will use $$T, N, B$$ for the Frenet frame of $$\alpha$$, and I will write the curvature and torsion of $$\alpha$$ as $$k$$ and $$\tau$$ without subscripts. Then by the structure equation, \begin{align} \alpha'' &= k N,\\ \alpha''' &= (kN)' = k'N + kN' = k'N + k(-kT + \tau B). \end{align}
Then $$\|\beta'\|=\|\alpha''\| = k$$. Also $$\|\beta'\times\beta''\| = \|\alpha''\times\alpha'''\| = \|kN\times (k'N+k(-kT+\tau B)\|= k^2\|k B + \tau T\| = k^2\sqrt{k^2+\tau^2},$$ by $$N\times N=0, N\times T=-B, N\times B=T$$.
Putting these together, we see $$k_\beta = \frac{\|\beta'\times\beta''\|}{\|\beta'\|^3} = \frac{k^2\sqrt{k^2+\tau^2}}{k^3} = \frac{\sqrt{k^2+\tau^2}}{k} = \sqrt{1+\frac{\tau_\alpha^2}{k_\alpha^2}}.$$